证明：连接$AO$并延长交$\odot O$于点$F$，连接$CF$，连接$BO$并延长，交$\odot O$与点$G$，连接$GE$，$AB$，$BE$，如图所示：

$\because PA$为$\odot O$的切线，
$\therefore FA\bot PA$，
$\therefore \angle PAF=90^{\circ}$，
$\therefore \angle KAP+\angle KAF=90^{\circ}$，
$\because AF$为直径，
$\therefore \angle ACF=90^{\circ}$，
$\therefore \angle ACE+\angle ECF=90^{\circ}$，
$\because \widehat {EF}=\widehat {EF}$，
$\therefore \angle KAF=\angle ECF$，
$\therefore \angle KAP=\angle ACE$，
$\because AC$∥$PB$，
$\therefore \angle KPE=\angle ACE$，
$\therefore \angle KPE=\angle KAP$，
$\because \angle PKE=\angle AKP$，
$\therefore \triangle KPE$∽$\triangle KAP$，
$\therefore \frac{{KP}}{{KA}}=\frac{{KE}}{{KP}}$，
$\therefore KP^{2}=KE\cdot KA$，
$\because PB$是$\odot O$的切线，
$\therefore \angle PBO=90^{\circ}$，
$\therefore \angle PBE+\angle EBG=90^{\circ}$，
$\because BG$为直径，
$\therefore \angle BEG=90^{\circ}$，
$\therefore \angle EBG+\angle EGB=90^{\circ}$，
$\therefore \angle PBE=\angle EGB$，
$\because \widehat {EB}=\widehat {EB}$，
$\therefore \angle EAB=\angle EGB$，
$\therefore \angle PBE=\angle EAB$，
$\because \angle BKE=\angle AKB$，
$\therefore \triangle BKE$∽$\triangle AKB$，
$\therefore \frac{{KB}}{{KA}}=\frac{{KE}}{{KB}}$，
$\therefore KB^{2}=KE\cdot KA$，
$\therefore KP=KB$，
$\because AC$∥$PB$，
$\therefore \triangle KPE$∽$\triangle ACE$，
$\therefore \frac{{PE}}{{CE}}=\frac{{KP}}{{AC}}$，
即$\frac{{PE}}{{CE}}=\frac{{KB}}{{AC}}$，
$\therefore PE\cdotAC=CE\cdotKB$.

证明：连接$AO$并延长交$\odot O$于点$F$，连接$CF$，连接$BO$并延长，交$\odot O$与点$G$，连接$GE$，$AB$，$BE$，如图所示：

$\because PA$为$\odot O$的切线，
$\therefore FA\bot PA$，
$\therefore \angle PAF=90^{\circ}$，
$\therefore \angle KAP+\angle KAF=90^{\circ}$，
$\because AF$为直径，
$\therefore \angle ACF=90^{\circ}$，
$\therefore \angle ACE+\angle ECF=90^{\circ}$，
$\because \widehat {EF}=\widehat {EF}$，
$\therefore \angle KAF=\angle ECF$，
$\therefore \angle KAP=\angle ACE$，
$\because AC$∥$PB$，
$\therefore \angle KPE=\angle ACE$，
$\therefore \angle KPE=\angle KAP$，
$\because \angle PKE=\angle AKP$，
$\therefore \triangle KPE$∽$\triangle KAP$，
$\therefore \frac{{KP}}{{KA}}=\frac{{KE}}{{KP}}$，
$\therefore KP^{2}=KE\cdot KA$，
$\because PB$是$\odot O$的切线，
$\therefore \angle PBO=90^{\circ}$，
$\therefore \angle PBE+\angle EBG=90^{\circ}$，
$\because BG$为直径，
$\therefore \angle BEG=90^{\circ}$，
$\therefore \angle EBG+\angle EGB=90^{\circ}$，
$\therefore \angle PBE=\angle EGB$，
$\because \widehat {EB}=\widehat {EB}$，
$\therefore \angle EAB=\angle EGB$，
$\therefore \angle PBE=\angle EAB$，
$\because \angle BKE=\angle AKB$，
$\therefore \triangle BKE$∽$\triangle AKB$，
$\therefore \frac{{KB}}{{KA}}=\frac{{KE}}{{KB}}$，
$\therefore KB^{2}=KE\cdot KA$，
$\therefore KP=KB$，
$\because AC$∥$PB$，
$\therefore \triangle KPE$∽$\triangle ACE$，
$\therefore \frac{{PE}}{{CE}}=\frac{{KP}}{{AC}}$，
即$\frac{{PE}}{{CE}}=\frac{{KB}}{{AC}}$，
$\therefore PE\cdotAC=CE\cdotKB$.